Let f : $R^n\to R$ be a differentiabile function in all points,with uniformly limited gradient .Prove that the gradient is continous in mean, that is: $\lim_{r\to 0} \frac{1}{\omega_n*r^n} \int_{B_r(x)} (\nabla f(y)-\nabla f(x)) dy=0$ ($\omega_n$ is the lebesgue measure of unitary ball $B_1(0)$ of $R^n$ )
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if the gradient is continous it's trivial but in general no. how can it be related with the uniformly limitated gradient – user62138 Aug 06 '13 at 15:56
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Uniformly bounded is the correct term. This assumption is made so that the gradient can be integrated; in general the gradient of a differentiable function can be non-integrable. We'll also need it in order to use the fundamental theorem of calculus in the proof: bounded gradient implies $f$ is Lipschitz, hence absolutely continuous on every line.
Fix $x$. By subtracting an appropriate linear function from $f$ we ensure that $f(x)=0$ and $\nabla f(x)=0$. Then $f(y)=o(|y-x|)$. Let's prove that $\int_{B_r(x)} f_1 =o(r^n)$ where $f_1$ is the derivative in the direction of $x_1$. Fix $x_2,\dots,x_n$ and integrate in the $x_1$ direction. The fundamental theorem of calculus says that the integral is the difference of two values of $f$, which is $o(r)$. Integration over the remaining $(n-1)$ coordinates yields $o(r)\omega_{n-1}r^{n-1}$, which is $o(r^n)$ as required.
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ok,but how can I prove with a counterexample that it doesn't work if we integrate $|\nabla f(y)-\nabla f(x)|$ – user62138 Aug 19 '13 at 18:21
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@user62138 Try $f(x)=x^2\sin (1/x)$ in one dimension. It's not easy to integrate $|f'|$ over $[-h,h]$ explicitly, but it's approximately $|\cos (1/x)|$, which averages to $2/\pi$. – user90090 Aug 19 '13 at 18:47